Since the average of the three integers will be 114/3 = 38, and the three numbers are consecutive, the numbers will be 36, 38 and 40.
Another way to do this problem using algebra is to let the first integer be n, then the next two are n+2 and n+4. Their sum is 3n+6 and it equals 114
So 3n+6=114 and
3n=108
so n=36
then next two numbers must be 38 and 40 since they are consecutive even integers.
The integers are 37, 38 and 39.
118
114/3 = 38, so the three integers are 37, 38 and 39.
27 + 28 + 29 + 30 = 114
Let the integers be x, (x+1) and (x+2) Since, x+(x+1)+(x+2) = -114, by solving the equation for x, we get that, x = -37 Hence the integers are -37, -36 and -35
The consecutive odd integers for 114 are 37, 38 and 39.
The integers are 37, 38 and 39.
118
The numbers are 56 and 58.
114/3 = 38, so the three integers are 37, 38 and 39.
The larger is 114, and the other is 112.
27 + 28 + 29 + 30 = 114
117
27 + 28 + 29 + 30 = 114
Let the integers be x, (x+1) and (x+2) Since, x+(x+1)+(x+2) = -114, by solving the equation for x, we get that, x = -37 Hence the integers are -37, -36 and -35
The integers are 27, 28, 29 and 30.
Three consecutive integers whose sum is 117 are 38, 39, and 40. N + (N+1) + (N+2) = 117 3N + 3 = 117 3N = 114 N = 38